Forming equations from situations

(a) $12$

1. Let the number be $n$.
   $$2n + 7 = 31$$
2. Solve.
   $$2n = 24 \Rightarrow n = 12$$

• M1 Forming $2n + 7 = 31$.
• M1 Reaching $2n = 24$.
• A1 The correct answer, $12$.

(a) $x + (x + 5) + 2(x + 5) = 4x + 15 = 83$

1. Write each person's stickers in terms of $x$: Tom has $x + 5$ and Ava has $2(x + 5)$.
2. Add the three amounts and set the total equal to 83.
   $$x + (x + 5) + 2(x + 5) = 83$$
3. Expand and collect like terms: $x + x + 5 + 2x + 10 = 4x + 15$, so $4x + 15 = 83$.

• P1 Writing Tom's amount as $x + 5$ and Ava's as $2(x + 5)$ (or $2x + 10$).
• P1 Adding all three expressions and setting the sum equal to 83.
• A1 Expanding and collecting to reach $4x + 15 = 83$ with every step shown.

(b) 44

1. Solve the equation: subtract 15, then divide by 4.
   $$4x = 68 \quad\Rightarrow\quad x = 17$$
2. Priya has 17, so Tom has $17 + 5 = 22$ and Ava has $2 \times 22 = 44$.

• M1 Solving to get $x = 17$ (or showing $4x = 68$).
• A1 Ava has 44 stickers.

(c) Yes: half of 83 is 41.5, and 44 is more than 41.5.

1. Half of all the stickers is $83 \div 2 = 41.5$.
2. Ava has 44, and $44 > 41.5$, so Ava is correct: she has more than half.

• M1 Finding half of the total, 41.5, or Ava's share of the total, $\frac{44}{83}$, or the other two's total, 39.
• C1 Saying yes, supported by a correct comparison: 44 is more than 41.5 (or 44 is more than 39).

(a) $5$ adult tickets

1. $$12a+7(9-a)=88$$
2. $$5a=25$$
3. Let a be the number of adult tickets, so there are 9 $-$ a child tickets.
4. The total cost gives 12a + 7(9 $-$ a) = 88.
5. Simplify to 5a + 63 = 88, so a = 5.

• P1 Establishing $12a+7(9-a)=88$ or an equivalent valid method.
• P1 Establishing $5a=25$ or an equivalent valid method.
• A1 Correct answer: $5$ adult tickets

(a) The square, by $4\ \text{cm}^2$

1. Write both perimeters in terms of $x$.
   $$\text{rectangle: } 2(3x - 2) + 2(x + 5) = 8x + 6, \qquad \text{square: } 4(x + 7) = 4x + 28$$
2. Set them equal and solve.
   $$8x + 6 = 4x + 28 \Rightarrow 4x = 22 \Rightarrow x = 5.5$$
3. Find each area.
   $$\text{rectangle: } 14.5 \times 10.5 = 152.25, \qquad \text{square: } 12.5^2 = 156.25$$
4. The square's area is greater, by $156.25 - 152.25 = 4\ \text{cm}^2$.

• P1 Writing both perimeters: $8x + 6$ and $4x + 28$.
• P1 Solving $8x + 6 = 4x + 28$ to get $x = 5.5$.
• P1 Working out both areas, $152.25$ and $156.25$.
• C1 Stating that the square is larger by $4\ \text{cm}^2$, from the two correct areas.

(a) $2$ m

1. $$(24-2w)(18-2w)=280$$
2. $$(w-2)(w-19)=0$$
3. Let the width be w metres. The central rectangle has dimensions 24 $-$ 2w and 18 $-$ 2w.
4. Its area gives (24 $-$ 2w)(18 $-$ 2w) = 280, so $4w^{2}$ $-$ 84w + 152 = 0.
5. Divide by 4 and factorise: $w^{2}$ $-$ 21w + 38 = (w $-$ 2)(w $-$ 19) = 0.
6. The width must be less than 9 m, so reject 19 and use w = 2 m.

• P1 Establishing $(24-2w)(18-2w)=280$ or an equivalent valid method.
• P1 Establishing $(w-2)(w-19)=0$ or an equivalent valid method.
• A1 Correct answer: $2$ m

(a) $8$ litres

1. $$2.4+0.45x=0.3(12+x)$$
2. $$0.15x=1.2$$
3. Let x litres be added. Concentrate volume becomes 2.4 + 0.45x.
4. For a 30% mixture, 2.4 + 0.45x = 0.30(12 + x).
5. Thus 0.15x = 1.2 and x = 8 litres.

• P1 Establishing $2.4+0.45x=0.3(12+x)$ or an equivalent valid method.
• P1 Establishing $0.15x=1.2$ or an equivalent valid method.
• A1 Correct answer: $8$ litres

(a) $9$ km/h

1. $$36/v+36/(v+3)=7$$
2. $$(7v+12)(v-9)=0$$
3. Let the outward speed be v km/h, with v > 0. Using time = distance/speed gives 36/v + 36/(v + 3) = 7.
4. Multiply by v(v + 3): 36(v + 3) + 36v = 7v(v + 3).
5. Rearrange: $7v^{2}$ $-$ 51v $-$ 108 = (7v + 12)(v $-$ 9) = 0.
6. The roots are $-12$/7 and 9. A speed is positive, so the outward speed is 9 km/h.

• P1 Establishing $36/v+36/(v+3)=7$ or an equivalent valid method.
• P1 Establishing $(7v+12)(v-9)=0$ or an equivalent valid method.
• A1 Correct answer: $9$ km/h

(a) $11$ rows

1. $$2a=3(n-1)$$
2. $$2an=330$$
3. $$(n-11)(n+10)=0$$
4. Let n be the number of rows and a the seats in the front row. The back row has a + 3(n $-$ 1) seats, so a + 3(n $-$ 1) = 3a and 2a = 3(n $-$ 1).
5. The total is $\frac{1}{2}n$(a + 3a) = 2an = 330.
6. Substitute 2a = 3(n $-$ 1): 3n(n $-$ 1) = 330, so $n^{2}$ $-$ n $-$ 110 = 0.
7. Factorise: (n $-$ 11)(n + 10) = 0. Reject $-10$ because the number of rows is positive.
8. There are 11 rows, with 15 seats in front and 45 at the back; $\frac{1}{2}$ $\times$ 11 $\times$ 60 = 330 checks the total.

• P1 Establishing $2a=3(n-1)$ or an equivalent valid method.
• P1 Establishing $2an=330$ or an equivalent valid method.
• P1 Establishing $(n-11)(n+10)=0$ or an equivalent valid method.
• A1 Correct answer: $11$ rows